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showing percentage in the result table

I wrote a sql statement in MS Access. the code is:

SELECT col1 as latency, count(col1) as repeat
FROM table1
group by col1
having (count(col1))>1
order by count(col1) desc

and result table is like

latency      repeat
1469      156
1453      154
1484      105
1468      85
1485      66
1500      66
1515      23

is it possible to have percentage in the 3rd column to decribe the percentage of 2nd column (repeat column). the result table looks like below

latency      repeat pct
1469      156   0.xx
1453      154    0.xx
1484      105    0.xx
1468      85      and so on
1485      66
1500      66
1515      23
0
rmtogether
Asked:
rmtogether
  • 2
2 Solutions
 
Raynard7Commented:
Select q.latency, q.repeat, (q.repeat / (select count(*) as cnt from table1 as t2)) as percentage_repeat
from
(
SELECT t1.col1 as latency, count(t1.col1) as repeat
FROM table1 as t1
group by t1.col1
having (count(t1.col1))>1
order by count(t1.col1) desc
) as q

Should give you a percentage that each repeat is of the whole volume - you just then need to format it (either in the query or report)
0
 
cdemirCommented:
hi,
in additioan to Raynard7's comment, i think there needs "sum(cnt)" because of "having (count(t1.col1))>1"

Select q.latency, q.repeat, (q.repeat / select sum(cnt) from (select count(*) as cnt from table1 as t2 group by col1 having count(*) > 1 )) as percentage_repeat
from
(
SELECT t1.col1 as latency, count(t1.col1) as repeat
FROM table1 as t1
group by t1.col1
having (count(t1.col1))>1
order by count(t1.col1) desc
) as q

0
 
cdemirCommented:
ops, missing paranthesis :

SELECT q.latency, q.repeat,
       (q.repeat / (SELECT SUM (cnt)
                      FROM (SELECT   COUNT (*) as cnt
                                FROM table1 as t2
                            GROUP BY col1
                              HAVING COUNT (*) > 1))) as percentage_repeat
  FROM (SELECT   t1.col1 as latency, COUNT (t1.col1) as repeat
            FROM table1 as t1
        GROUP BY t1.col1
          HAVING (COUNT (t1.col1)) > 1
        ORDER BY COUNT (t1.col1) DESC) as q
0
 
rmtogetherAuthor Commented:

thanks guys!

i think cdemir is right I need  "sum(cnt)".
0

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